bibkey: stadlmann2022meansquaregaps authors: Julia Stadlmann year: 2022 title: On the mean square gap between primes doi: null url: https://arxiv.org/abs/2212.10867v1 claim: Theorem 1 gives an unconditional prime-gap square sum. Applications bound actual CA price residence, quantify the selected-source state count and transport its missing upper bound to the existing prime-prefix surplus; no signed surplus count estimate is supplied. strata_touched: [] license: citation-only triage: anchor

Prime-gap moments and actual CA residence lengths

Julia Stadlmann, On the mean square gap between primes, arXiv:2212.10867v1, Theorem 1, PDF p.1, states unconditionally that, for every fixed $ϵ>0$,

$$\sum _{p_n\le x}(p_{n+1}-p_n{)}^2{\ll }_{ϵ}{x}^{123/100+ϵ}.\text{\hspace{1em}(R1)}$$

The versioned HTML and PDF supply the primary statement. This note uses that theorem without re-proving or independently auditing its proof, and makes no claim that its exponent is the best available in all later literature. The following is an application to the project’s existing actual CA path, not a theorem stated by Stadlmann, an originality claim, or Lean certification.

Transport the existing gap moment to the actual price clock

Use the existing full CA optimizer $C_b$ at $g(b)=1/(b\log b)$, choosing the largest optimizer and retaining all activation ties and repeated prime-power layers. Put $B(b)=\log C_b$; it is generally not $b$. No regularity or GA1 condition is imposed on subsequent states.

The existing first-layer activation rule places the event for prime $p$ at the unique $\tau (p)>1$ with

$$g(\tau (p))=\frac{\log (1+1/p)}{\log p}.$$

The already used bounds $1/(p+1)<\log (1+1/p)<1/p$, together with decreasing $g$, give $p<\tau (p)<p+1$. Consequently the first-layer parent cell has length

$$\tau (p_{n+1})-\tau (p_n)\le p_{n+1}-p_n+1\le 2(p_{n+1}-p_n).$$

For sufficiently large $X$, let $J_j$ be the complete positive-length intervals on which $C_b$ is constant and whose intersection with $[X,3X]$ has positive length. Write ${\ell }_j=|J_j|$ for their full lengths, including any portion outside that observation interval. Every $J_j$ lies in one first-layer parent cell. All higher layers merely subdivide that cell; simultaneous ties add no positive-length interval. Within each parent cell, the sum of the squared lengths of any subcollection is at most the square of the parent’s length. Every relevant parent’s left event is at most $3X$, hence its left prime is less than $3X$. Applying (R1) at $3X$ therefore gives

$$\boxed{\sum _j{\ell }_j^2{\ll }_{ϵ}{X}^{123/100+ϵ}.}\text{\hspace{1em}(R2)}$$

The original theorem includes the gap whose left prime is below the cutoff even if its right prime is above it. Thus (R2) pays both boundary-straddling cells without dropping any full length. Endpoint conventions have zero price measure.

For any collection $\mathcal{J}$ of $K$ such plateaus, Cauchy’s inequality gives the actual-price measure bound

$$\boxed{\left|[X,3X]\cap \bigcup _{j\in \mathcal{J}}{J}_j\right|{\ll }_{ϵ}{X}^{123/200+ϵ/2}\sqrt{K}.}\text{\hspace{1em}(R3)}$$

The collection may depend on the actual prime data; no independence assumption is made. Likewise the price measure occupied by full plateaus longer than $\sqrt{X}$ satisfies

$$\left|\{b\in [X,3X]:|J(b)|>\sqrt{X}\}\right|{\ll }_{ϵ}{X}^{73/100+ϵ}.\text{\hspace{1em}(R4)}$$

Indeed, each such length obeys ${\ell }_j\le {\ell }_j^2/\sqrt{X}$. Choosing the theorem’s free parameter $ϵ=1/100$ gives $O(X^{74/100})$, smaller than $X^{1-{\eta }^{\prime }/2}\sqrt{\log X}$ for each fixed $0<{\eta }^{\prime }<1/2$. This does not discard those plateaus from a signed mean: their excess amplitudes would still need a bound.

The remaining count must retain the Robin sign

Use the particular fixed ${\eta }^{\prime }\in (0,1/2)$, $c>0$ and $\delta =c/(42^{{\eta }^{\prime }})$ from the existing selected-source persistence application. Let $K_{{\eta }^{\prime },\delta }(X)$ count the plateaus above for which $\log G(C_b)-\gamma >\delta X^{-{\eta }^{\prime }}$. The quantity is constant on each plateau. An actual selected source with $A=\log N\in [X,2X]$ supplies an interval within $[X,3X]$ of length at least

$$\lambda X^{1-{\eta }^{\prime }/2}\exp \left(\frac{(\log X{)}^{1/4}}{2}\right)$$

entirely above that threshold. Combining its length with (R3) forces, on any such scale,

$$K_{{\eta }^{\prime },\delta }(X){\gg }_{ϵ,\lambda }{X}^{77/100-{\eta }^{\prime }-ϵ}\exp \left((\log X{)}^{1/4}\right).\text{\hspace{1em}(R5)}$$

The sufficient arithmetic count bound would therefore be $o!\left(X^{77/100-\eta’-\epsilon} e^{(\log X)^{1/4}}\right)$ for any one fixed $0<ϵ<27/100$. No such sign-sensitive count or directed mean bound is supplied. The unconditional input (R2) controls residence lengths; the existing persistence supplies amplitude only when a selected excess source already exists. Neither controls how often actual CA states exceed the threshold. The complete original Robin tail and RH remain unproved.

The existing uniform prime count gives a stronger source-specific count

For these selected persistence intervals, (R5) is a weaker count than the one supplied directly by the existing uniform extension (GS4). Reuse that estimate and the same source’s (P8), keeping $h=\lambda A^{1-{\eta }^{\prime }/2}{e}^{(\log A{)}^{1/4}/2}$. The following application neither re-proves a prime-count theorem nor supplies a sign-sensitive count upper bound.

Every prime in $(A+1,A+h-1]$ has its first-layer event strictly inside $(A,A+h)$. The event map $\tau (p)$ is strictly increasing: $\log (1+1/p)$ decreases and $\log p$ increases with $p$, so its activation price decreases strictly. Each event introduces a new prime and therefore a different actual CA integer. Higher-layer ties do not erase this distinction. Local finiteness gives the largest-tie post-event state a positive-length plateau intersecting the persistence interval, entirely above its threshold there. Thus all of these first-layer events count toward $K_{{\eta }^{\prime },\delta }(X)$.

For each particular fixed $0<{\eta }^{\prime }<1/2$, $h-2\gg A^{2/3}$ and $h=o(A)$. Apply (GS4) at $Y=2A$ to $[A+1,A+h-1]$. Its already paid prime-power correction is $o(h)$, giving a $\vartheta$ increment $h(1+o(1))$. All prime weights in that interval are uniformly $(\log A)(1+o(1))$, hence the existing estimate gives $\pi (A+h-1)-\pi (A+1)=(1+o(1))h/\log A$. Consequently every scale carrying the same selected source obeys

$$\boxed{K_{{\eta }^{\prime },\delta }(X){\gg }_{\lambda ,{\eta }^{\prime }}\frac{X^{1-{\eta }^{\prime }/2}}{\log X}\exp \left(\frac{(\log X{)}^{1/4}}{2}\right).}\text{\hspace{1em}(R6)}$$

The asymptotic threshold may depend on the fixed exponent; it is not uniform as ${\eta }^{\prime }\downarrow 0$. The ratio of (R6)’s lower scale to (R5)’s is $X^{23/100+{\eta }^{\prime }/2+ϵ}{e}^{-(\log X{)}^{1/4}/2}/\log X$, which tends to infinity. Thus an arithmetic count upper bound $o\left(X^{1-{\eta }^{\prime }/2}{e}^{(\log X{)}^{1/4}/2}/\log X\right)$ would already exclude the selected sources. Requiring the smaller upper bound after (R5) is unnecessary for that task.

No such upper bound is established. The unrestricted residence comparison (R2)–(R4) still applies to arbitrary data-dependent state collections; the direct first-prime count applies to the already justified persistence interval. Both preserve the same actual source, and neither pays the complete signed Robin tail or proves RH.

Count upper bounds can reuse the existing prime-prefix surplus

Keep the full largest-optimizer path and the positive-length-state count $K_{{\eta }^{\prime },\delta }(X)$ from (R6). The existing Nicolas prime-prefix readout can be used at the actual support of each CA state. For a prime $p\ge 5$ define

$$\mathcal{P}(p)=\prod _{q\le p}(1-1/q{)}^{-1},U(p)=\log \mathcal{P}(p)-\gamma -\log \log \vartheta (p).$$

This is the existing $-\log f(p)$; it is not a new prime-error criterion or a proved estimate. Here $p$ is the largest prime actually dividing the state, not the size-envelope cutoff for that integer. The bound uses its radical and keeps that same integer.

For an actual state $n={\prod }_{q\le p}{q}^{a_q}$ with every $a_q\ge 1$ and $B=\log n$, the usual Euler formula gives

$$\begin{array}{rl}\log G(n)-\gamma =& U(p)+\sum _{q\le p}\log (1-q^{-a_q-1})\\ & +\log \log \vartheta (p)-\log \log B\le U(p).\end{array}\text{\hspace{1em}(R7)}$$

Every exponent remains in the displayed identity. Both correction terms are nonpositive because $B\ge \vartheta (p)>1$. No assumption about independent prime errors, favorable layer timing, or $B=b$ enters this comparison.

Directly reuse the existing higher-layer event count from the FIB theory volume, §420, equation (420.5). With the same activation index $j\ge 2$, its instance at $T=3X$ is

$$H(X):=E_{\ge 2}(3X)=\#\{(q,j):j\ge 2,{\tau }_{q,j}\le 3X\}=O(\sqrt{X}(\log X{)}^{3/2}).\text{\hspace{1em}(R8)}$$

This counts each layer in a tied cluster separately and includes layers already active at the left endpoint. The existing proof and its activation cutoff are not repeated here.

There is at most one plateau crossing the left boundary. Each remaining plateau counted by $K$ begins at an event in $(X,3X)$. If that event contains a first layer, its new prime $p$ is the largest active prime, its post-event state satisfies (R7), and $X-1<p<3X$ by $p<\tau (p)<p+1$. Different first-layer events have different primes. If it contains no first layer, charge the new state to one of its higher-layer pairs. A simultaneous cluster creates one post-event state; no isolated partial tied optimizer is inserted into the path. Therefore

$$\boxed{K_{{\eta }^{\prime },\delta }(X)\le 1+H(X)+\#\{p\text{ prime}:X-1<p<3X,U(p)>\delta X^{-{\eta }^{\prime }}\}.}\text{\hspace{1em}(R9)}$$

For each particular fixed $0<{\eta }^{\prime }<1/2$, (R8) is little-o of the lower-count scale in (R6). Thus a prime-prefix surplus count little-o of that scale would provide the missing upper count. No such signed prime count is proved. This application retains all states and exponents while paying their additional count explicitly; it neither upper-bounds the count by a lower selector of all-positive cells nor relies on lower bounds for plateau lengths. The existing Nicolas formula already relates $U(p)$ to its full signed integral and endpoint correction. No new prime-distribution theorem, originality claim, Lean certification, full Robin tail estimate or RH proof is supplied here.